\newproblem{lay:6_5_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.5.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $m\times n$ matrix such that $A^TA$ is invertible. Show that the columns of $A$ are linerly independent.
	[\textit{Careful}: You may not assume that $A$ is invertible; it may not even be square.] 
}{
   % Solution
	By the Invertible Matrix Theorem (see Section 2.9), if $A^TA$ is invertible, then $\mathrm{Nul}\{A^TA\}=\mathbf{0}$, that means
	that the only solution of the problem
	\begin{center}
		$A^TA\mathbf{x}=\mathbf{0}$
	\end{center}
	is the vector $\mathbf{x}=\mathbf{0}$. In Exercise 6.5.19 we showed that $\mathrm{Nul}\{A^TA\}=\mathrm{Nul}\{A\}$, so the only 
	vector in $\mathrm{Nul}\{A\}=\{\mathbf{0}\}$, that is the only solution of the problem
	\begin{center}
		$A\mathbf{x}=\mathbf{0}$
	\end{center}
	is $\mathbf{x}=\mathbf{0}$, and consequently, the columns of $A$ are linearly independent.
}
\useproblem{lay:6_5_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
